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I see this notation a lot but I am not sure why we write $\int_{0}^tf(r)dr$ instead of $\int_0^tf(t)dt$. If we assume $\int f(x)dx=F(x)$ we have:

$$\int_0^t f(r)dr=F(r)|_{r=0}^{r=t}=F(t)-F(0)$$ $$\int_0^t f(t)dt=F(t)|_{t=0}^{t=t}=F(t)-F(0)$$

And we get $F(t)-F(0)$ in both case.I saw this notation in such integrals a lot but I don't understand the logic of presenting a new variable (which is "$r$" here) to calculate the definite integral. And is it logically (or mathematically) wrong to write $\int_0^tf(t)dt$ instead of $\int_0^tf(r)dr$?

Etemon
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    Yes, $\int_0^{t}f(t)dt$ does not make sense. – Kavi Rama Murthy Oct 09 '20 at 11:29
  • @KaviRamaMurthy I don't know the reason. It seems $\int_0^{t}f(t)dt$ means the area under the curve $f(t)$ where $t\in (0,t)$. and it seems logical to me. So I don't completely understand why it doesn't make sence. – Etemon Oct 09 '20 at 11:30
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    You are using the same variable twice as a variable in $f(t)$, and as a number in the bounds of integration from $0$ to $t$. $t$ cannot be a variable and a number at the same time. – Toby Mak Oct 09 '20 at 11:32

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